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Calculating Current using Kirkoff's Rules

  1. Oct 13, 2012 #1
    1. The problem statement, all variables and given/known data
    Apply the loop rule to two of the three loops to calculate I1,I2,and I3.

    Picture of DC circuit involved
    cVrkF.jpg

    2. Relevant equations

    V=IR

    3. The attempt at a solution
    L1=2v+2I2-4I1=0 (simplified)
    and
    L3=4-4I1-5I3=0

    But if you work through the algebra you'd get
    I2=1
    which would then make
    I1=1

    which does not work for the junction rule of
    I3=I1-I2 cause that is 0.

    The rules still confuse me when look at some of the resistors on the right side I can tell which way current flows, obviously did it wrong though, any help would be appreciated.
     
  2. jcsd
  3. Oct 13, 2012 #2

    Attached Files:

  4. Oct 13, 2012 #3

    OmCheeto

    User Avatar
    Gold Member
    2016 Award



    1Ω + 3Ω ≠ 2Ω
     
  5. Oct 13, 2012 #4
    Okay, re-attempted using loop one and loop two, does, this seem more accurate?

    L1=12V-I1(2)+I2(1)-10V+I2(3)-I1(1)=0
    =>
    L1=2V-4I1+4I2=0


    L2=-3(I2)+10V-I2(1)-I3(1)-8V-I3(2)=0
    =>
    L2=2V-4I2-5I3=0
    =>
    L2=2-4I2-5(I1-I2)=0
    =>
    L2=2-4I2-5I1+5I2=0
    =>
    L2=2-5I1+I2=0

    So from there
    I2=5I1-2


    Finally plugging it all through
    I1=I2+(1/2)[From the first loop equation]
    =>
    I1=5I1-2+(1/2)
    =>
    I1=(3/8)

    if thats true

    I2=5(3/8)-2=(-1/8) [Is a negative current okay???]

    and from there

    I3=I1-I2=(3/8)-(-1/8)=1/2

    The negative current for I2 feels wrong, did I mess up any of the loop equation or any algebra along the way?
     
  6. Oct 13, 2012 #5
    For the second loop, it should be +5I2, not -5I2
    As for there being a negative current, it just means it's flowing in the opposite direction from what you set it up as in the problem.
     
  7. Oct 14, 2012 #6
    Okay, think I got it

    You're right it should be +5I3 because it's with the current, just like I2 works out in the first loop equation

    L1=2-4I1+4I2=0

    Using this to solve for I1

    I1=I2+(1/2)

    Now I did this wrong, so going back through the derivation of this loop equation

    L2=-3I2+10-I2+2I3+I3-8+2I3=0
    =>
    2-4I2+5I3=0
    =>
    2-4I2+5(I1-I2)=0
    =>
    L2=1+5I1-9I2=0


    Using that to solve for I2

    I2=(5/9)I1+(2/9)

    Solving for the numerical value of I1
    I1=((5/9)I1+(2/9))+(1/2)
    =>
    I1=(5/9)I1+(13/18)
    =>
    (4/9)I1=(13/18)
    =>
    I1=(13/8)


    Solving for the numerical Value of I2

    I2=(5/9)(13/8)+(2/9)
    =>
    I2=(9/8)


    Which Finally implies

    I3=I1-I2
    =>
    I3=(13/8)-(9/8)=(1/2)

    Aye?Nay?
     
  8. Oct 15, 2012 #7
    correct
    quick note: you had I1=I2+1/2, so you could have easily figured out I3 right then and there. I3 = I1-I2 = I2+1/2-I2 = 1/2
     
  9. Oct 15, 2012 #8

    OmCheeto

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    2016 Award

    really?



    L2=1+5I1-9I2=0
    i1=13/8
    i2=9/8
    ergo:
    1+5*(13/8)-9*(9/8)=0
    8/8+65/8-81/8=0
    73/8-81/8=0
    -8/8=0
    -1=0 ?

    Don't all 4 equations have to come out correctly?
     
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